China Girls' Mathematical Olympiad

For every $i$ with $1\le i\le n$, we define set $$S_i=\{b\mid b\text{ is an element in }S\text{ and is divisible by }{a}_i\}.$$ There are $\lfloor \frac{m}{a_i}\rfloor$ elements in $S_i$. Since every element in $S$ is not divisible by any two distinct elements in $T$, it follows that $S_i\cap S_j=\varnothing$ for $1\le i<j\le n$. Thus $$\sum _{i=1}^n\lfloor \frac{m}{a_i}\rfloor =\sum _{i=1}^n|S_i|\le |S|=m.$$ Note that $\frac{m}{a_i}<\lfloor \frac{m}{a_i}\rfloor +1$. It follows that $$\sum _{i=1}^n\frac{m}{a_i}<\sum _{i=1}^n(\lfloor \frac{m}{a_i}\rfloor +1)=\sum _{i=1}^n\lfloor \frac{m}{a_i}\rfloor +\sum _{i=1}^{n}1\le m+n.$$ So the desired result is obtained.